tag:blogger.com,1999:blog-6933544261975483399.post4160780931987057257..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1310 Square, Center, Right Triangle, Area, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6933544261975483399.post-79935478211716030742018-03-19T00:11:28.410-07:002018-03-19T00:11:28.410-07:00Genius. Genius. SSCMATHhttps://www.blogger.com/profile/14414915087362505064noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-2848044817824196562017-01-29T00:59:21.463-08:002017-01-29T00:59:21.463-08:00AC = a \/2 (a sqrt2)AC = a \/2 (a sqrt2)c.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-288764479863941162017-01-28T22:24:05.573-08:002017-01-28T22:24:05.573-08:00Excellent Excellent Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-40778662286994042452017-01-28T17:46:52.887-08:002017-01-28T17:46:52.887-08:00Note that AF = 15/sqrt2
Using Ptolemy in AOBC
sq...Note that AF = 15/sqrt2<br /><br />Using Ptolemy in AOBC<br /><br />sqrt2.a.x + a.BC = AB.a<br /><br />Now note that BC = sqrt2. a(a-b)/15<br /><br />and AB = 15/sqrt2 + b.(a-b)/sqrt2/15<br /><br />Now substitute these in the Ptolemy equation Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-84404154118606248002017-01-28T16:26:34.880-08:002017-01-28T16:26:34.880-08:00Full Ptolemy isn't really necc. once you deriv...Full Ptolemy isn't really necc. once you derive AF = 15/sqrt(2). Then since its cyclic OBF is similar to ACF and x / AC = OF / AF or x = AC * OF / AF. <br /><br />AC = sqrt(2) * a since its a 45-45-90 and OF = 30 / a using the AFO's area.<br /><br />So AC * OF = 30 * sqrt(2) and 30 * sqrt(2) / (15 / sqrt(2)) = 4.Benjamin Leishttps://www.blogger.com/profile/10974191081762367425noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-4039852089544179872017-01-28T12:14:03.068-08:002017-01-28T12:14:03.068-08:00To Sumith
On what relation is included xTo Sumith<br />On what relation is included xc.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-5607221238523228792017-01-27T18:52:49.149-08:002017-01-27T18:52:49.149-08:00Let AO = a and OF = b so that FC = a-b and AC = sq...Let AO = a and OF = b so that FC = a-b and AC = sqrt2.a<br /><br />ab = 30......(1)<br /><br />From areas of similar triangles<br /><br />(a^2+b^2)/(a-b)^2 = 15/7 which simplifies to <br />a^2+b^2 = 225/2 ...,(2) using (1)<br /><br />Now use Ptolemy in cyclic quadrilateral AOBC, noting that BC = (a-b)a/sqrt(a^2+b^2) using Pythagoras and we get after much simplification <br /><br />x = 2ab/sqrt2(a^2+b^2) = 60/sqrt225 (from (2)) = 4<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.com